Q54 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

Assertion (A) If $a_1, a_2, \ldots, a_n$ are the $n$ distinct roots of the equation $x^n-2=0$, then $1+\left(1-a_1\right)\left(1-a_2\right) \ldots\left(1-a_{n-1}\right)\left(1-a_n\right)=0$ Reason (R) If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are the roots of $f(x) \equiv p_0 x^n+p_1 x^{n-1}+p_2 x^{n-2}+\ldots+p_n=0$, then the roots of $f(g(x))=0 \operatorname{are~}^{-1}\left(\alpha_i\right), i=1,2,3, \ldots, n$ The correct option among the following is

  • Correct answer
    A
    (A) is true, (R) is true and (R) is the correct explanation for (A).
  • B
    (A) is true, (R) is true but (R) is not the correct explanation for (A).
  • C
    (A) is true but (R) is false .
  • D
    (A) is false but (R) is true.

Solution

Given, $a_1, a_2, a_3, \ldots, a_n$ are $n$ distinct roots of $\begin{aligned} & x^n-2=0 \\ & \therefore x^n-2=\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_{n-1}\right)\left(x-a_n\right) \\ & \text { put } x=1 \\ & \quad 1-2=\left(1-a_1\right)\left(1-a_2\right) \ldots\left(1-a_{n-1}\right)\left(1-a_n\right) \\ & \Rightarrow \quad 1+\left(1-a_1\right)\left(1-a_2\right) \ldots\left(1-a_{n-1}\right)\left(1-a_n\right)=0 \\ & \therefore A \text { is true } \\ & \text { If } \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_n \text { are } n \text { root of } f(n) \\ & \therefore \quad f\left(\alpha_1\right)=f\left(\alpha_2\right)=\ldots=f\left(\alpha_n\right)=0 \\ & \text { If } \alpha_i=g(x) \Rightarrow x=g^{-1}\left(\alpha_i\right) \end{aligned}$ Then root of $f(x)$ is $ f g(x)=0 \text { are } g^{-1}\left(\alpha_i\right), i=1,2,3 \ldots $ $R$ is also true and correct explanation for $A$.

Hence, the correct answer is option: 'A'.

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